Question

In basic medium, find equivalent weight.
$N{a}_2{S}_2{O}_3\to N{a}_{2}S{O}_4$.

Answer 1

$N{a}_2{S}_2{O}_3\to N{a}_{2}S{O}_4$

Balanced $R{x}^n$

$N{a}_2{S}_2{O}_3+2O{H}^{-}\to 2N{a}^{+}+2S{O}_4^{2-}+H_{2}O$

Molar mass of $N{a}_2{S}_2{O}_3=\frac{23236448}{158}$

Equivalent weight $=\frac{MolarmassofN{a}_2{S}_2{O}_3}{eqof{H}^{+}}$

$=\frac{158}{2}=79g/eq.of{H}^{+}$

$\therefore$ one mole of $N{a}_2{S}_2{O}_3$ reacts with two equivalent of $H^{+}$